Question 1051583: Can someone help me with this problem:
In a certain production process, 25% of the items produced are found defective. It therefore requires through inspection of all the items produced before packing them for sale.
a) Develop a probability density function for the number of non-defective items in a day's production of n items.
b) Using your answer to a), determine the probability of getting at least 50 good items if 100 items are produced.
Found 2 solutions by rothauserc, ewatrrr:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We use the binary probability distribution
:
The probability of non-defective items is 1 - 0.25 = 0.75
:
Probability(k items are non-defective) = nCk * p^k * q^(n-k), where n is number of items produced, k = number of non-defective items, p = probability of non-defective items, q = 1-p, nCk of the combination of n items taken k at a time
:
a) Probability(k items are non-defective) = nCk * (0.75)^k * (0.25)^(n-k)
:
b) Probability(at least 50 items are produced out of 100) is the summation of
the probabilities for 50, 51, 52, ..., 100
:
Probability(at least 50 items are produced out of 100) = 0.999999
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website! a)
p(def) = .25, p(not def) = .75
probability density function for the number of non-defective items in a day's production of n items.
Let x represent the number of non-defective itemsin a day's production of n items
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b)
p = .75, n = 100
P (x>= 50) Commulative Probability
P (x>= 50) = 1 - binomcdf(100,.75,49) = 1 - .000002 = .999998
P (x>= 50) basically a sure thing.
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